Study notes

Indices & Standard Form

The laws of indices, including negative and fractional indices, and writing/calculating with numbers in standard form.

Learn it step by step

The three basic laws of indices work with the SAME base

To multiply powers of the same base, add the indices: $a^{m} \times a^{n} = a^{m + n}$ . To divide, subtract: $a^{m} \div a^{n} = a^{m - n}$ . To raise a power to a power, multiply: $(a^{m})^{n} = a^{mn}$ . These only apply when the base $a$ is the same on both sides.

A zero index is always 1; a negative index means 'reciprocal'

Any nonzero base raised to the power 0 equals 1: $a^{0} = 1$ . A negative index does not make the value negative, it flips it: $a^{- n} = \frac{1}{a ^{n}}$ . For example $5^{- 2} = \frac{1}{5 ^{2}} = \frac{1}{25}$ , not $- 25$ .

A fractional index is a root, possibly followed by a power

$a^{1/ n} = n a$ (the $n$ th root). More generally $a^{m / n} = (n a)^{m}$ : take the $n$ th root first, then raise the result to the power $m$ . Doing the root first keeps the numbers small, e.g. $8^{2/3} = (38)^{2} = 2^{2} = 4$ , not $3 8^{2} = 364$ worked the hard way (same answer, but harder arithmetic).

Negative AND fractional together: deal with them one at a time

For $a^{- m / n}$ , first take the root and power (the fractional part), then take the reciprocal (the negative part), or vice versa, in either order. $1 6^{- 3/4} = \frac{1}{1 6 ^{3/4}} = \frac{1}{( 4 16 ) ^{3}} = \frac{1}{2 ^{3}} = \frac{1}{8}$ .

Standard form writes very large or very small numbers as $a \times 10^n$

Standard form is $a \times 1 0^{n}$ where $1 \leq a < 10$ and $n$ is an integer (positive for large numbers, negative for numbers less than 1). $4500000 = 4.5 \times 1 0^{6}$ ; $0.00072 = 7.2 \times 1 0^{- 4}$ .

Multiplying or dividing in standard form: combine coefficients and add/subtract powers, then renormalise

Multiply (or divide) the coefficients together, and add (or subtract) the powers of 10 using the index laws. If the resulting coefficient is not between 1 and 10, shift the decimal point and adjust the power of 10 so the answer is properly in standard form.

Worked examples

Simplify

(2 x^{2})^{3} \div (4 x^{5})

, giving your answer in the form

k x^{n}

Cube the bracket first: $(2 x^{2})^{3} = 2^{3} x^{6} = 8 x^{6}$ .
Divide by $4 x^{5}$ : $8 \div 4 = 2$ , and $x^{6} \div x^{5} = x^{6 - 5} = x^{1}$ .
Answer: $2 x$ .

Evaluate

3 2^{2/5}

Take the 5th root first: $532 = 2$ (since $2^{5} = 32$ ).
Raise the result to the power 2: $2^{2} = 4$ .
$3 2^{2/5} = 4$ .

Evaluate

4 9^{- 1/2}

Deal with the fractional part first: $4 9^{1/2} = 49 = 7$ .
The negative sign means take the reciprocal: $4 9^{- 1/2} = \frac{1}{7}$ .
Forgetting the negative sign would incorrectly give 7 instead of $\frac{1}{7}$ .

Write

0.000063

in standard form, then calculate

(6.3 \times 1 0^{- 5}) \times (2 \times 1 0^{3})

, giving the final answer in standard form.

$0.000063 = 6.3 \times 1 0^{- 5}$ (the decimal point moves 5 places to make $1 \leq 6.3 < 10$ ).
Multiply coefficients: $6.3 \times 2 = 12.6$ . Add powers of 10: $1 0^{- 5 + 3} = 1 0^{- 2}$ .
This gives $12.6 \times 1 0^{- 2}$ , but $12.6$ is not between 1 and 10, so renormalise: $12.6 \times 1 0^{- 2} = 1.26 \times 1 0^{- 1}$ .

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